前缀和&差分

两个常见的互逆运算

一维前缀和

顾名思义，前缀和就是一个数组前缀的和说了和没说一样

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#include <iostream>
using namespace std;
int N, A[10000], B[10000];
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  // 前缀和数组的第一项和原数组的第一项是相等的。
  B[0] = A[0];
  for (int i = 1; i < N; i++) {
    // 前缀和数组的第 i 项 = 原数组的 0 到 i-1 项的和 + 原数组的第 i 项。
    B[i] = B[i - 1] + A[i];
  }
  for (int i = 0; i < N; i++) {
    cout << B[i] << " ";
  }
  return 0;
}

有手就行的求法。。。

二维前缀和

定义 sumx,y=∑i=1x∑j=1yai,jsum_{x,y}=\sum_{i=1}^{x} \sum_{j=1}^y a_{i,j}

则您可以先手算一

请先手推

sumi,j=sumi−1,j+sumi,j−1−sumi−1,j−1+ai,jsum_{i,j}=sum{i-1,j}+sum_{i,j-1}-sum_{i-1,j-1}+a_{i,j}

则(x1,y1)−(x2,y2)(x_1,y_1)-(x_2,y_2)这个区间的合为

请先手推

sumx2,y2−sumx1,y2−sumx2,y1−1+sumx1−1,y1−1sum{x_2,y_2}-sum_{x_1,y_2}-sum_{x_2,y_1-1}+sum_{x_1-1,y_1-1}

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#include <algorithm>
#include <iostream>
using namespace std;
int a[103][103];
int b[103][103];  // 前缀和数组，相当于上文的 sum[]
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      cin >> a[i][j];
      b[i][j] =
          b[i][j - 1] + b[i - 1][j] - b[i - 1][j - 1] + a[i][j];  // 求前缀和
    }
  }
  int ans = 1;
  int l = 2;
  while (l <= min(n, m)) {  // 判断条件
    for (int i = l; i <= n; i++) {
      for (int j = l; j <= m; j++) {
        if (b[i][j] - b[i - l][j] - b[i][j - l] + b[i - l][j - l] == l * l) {
          ans = max(ans, l);  // 在这里统计答案
        }
      }
    }
    l++;
  }
  cout << ans << endl;
  return 0;
}

高维前缀和

似乎已经不是普及了
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for(int j = 0; j < n; j++) 
    for(int i = 0; i < 1 << n; i++)
        if(i >> j & 1) f[i] += f[i ^ (1 << j)];

差分

对于数组a[1...n]a[1...n]，要使得区间[L…R]统一+x+x则我们可以在差分数组d[1..n]d[1..n]中 dl+=xd_{l}+=x 并在 dr+1−=xd_{r+1}-=x

例题 P2367 语文成绩

很明显的差分板子题

Code:

建议读者自己先Code
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// Problem: P2367 语文成绩
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P2367
// Memory Limit: 125 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 5e6 + 10;

// bool st;
int n, m;
int l, r, x;
int a[N];
int d[N];
int res = INF;
int now = 0;
// bool en;

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= m; i++)
    {
        cin >> l >> r >> x;
        d[l] += x;
        d[r + 1] -= x;
    }
    for (int i = 1; i <= n; i++)
    {
        now += d[i];
        res = min(res, a[i] + now);
    }
    cout << res << endl;
    return 0;
}